# Math Help - Supremum property: Inequality

1. ## Supremum property: Inequality

Hello everybody,
I do not understand a part of a proof where the author makes the following observation for given $x,y \in R$:

$\sup\{b^s b^{r-s}|r\le x+y, s\le x, r,s \in Q\} \ge \sup\{\sup B(x) b^{r-s}|r\le x+y, s\le x, r,s \in Q\}$

with:

$B(x) = \{b^s|s\le x, s \in Q\}$

I thought that it would be exactly the other way round so that the left side of the inequality was smaller. Is the supremum as an upper bound not always bigger then any element $b^s$ of the set? Therefore the supremum of the set where the elements are multiplied with this supremum should be bigger than the supremum of the set where the same elements are only multiplied with a (lower or equal) value of the same set. I think I am misinterpreting something and I wonder if anybody could help me here since I am stuck for a longer while now. The full proof I am looking at can be found here: http://www.csie.ntu.edu.tw/~b89089/b...in/Rudin_1.pdf (It's number 6d)

Thank you all! Best, Rafael

2. Originally Posted by raphw
Hello everybody,
I do not understand a part of a proof where the author makes the following observation for given $x,y \in R$:

$\sup\{b^s b^{r-s}|r\le x+y, s\le x, r,s \in Q\} \ge \sup\{\sup B(x) b^{r-s}|r\le x+y, s\le x, r,s \in Q\}$

with:

$B(x) = \{b^s|s\le x, s \in Q\}$

I thought that it would be exactly the other way round so that the left side of the inequality was smaller. Is the supremum as an upper bound not always bigger then any element $b^s$ of the set? Therefore the supremum of the set where the elements are multiplied with this supremum should be bigger than the supremum of the set where the same elements are only multiplied with a (lower or equal) value of the same set. I think I am misinterpreting something and I wonder if anybody could help me here since I am stuck for a longer while now. The full proof I am looking at can be found here: http://www.csie.ntu.edu.tw/~b89089/book/Rudin/Rudin_1.pdf (It's number 6d)

Thank you all! Best, Rafael
The inequality is false, because
$sup\{b^s b^{r-s}|r\le x+y, s\le x, r,s \in Q\}=b^{x+y}$
but the set
$\{\sup B(x) b^{r-s}|r\le x+y, s\le x, r,s \in Q\}$
is actually unbounded and therefore has no supremum.
For r=x+y and $s\rightarrow-\infty$
$b^{r-s}=b^{(x+y)-s}\rightarrow\infty (since b>1)
$

Correctly it should be
$sup\{b^s b^{r-s}|r\le x+y, s\le x, r,s \in Q\} \geq \sup\{\sup B(x) b^{t}| t\leq y , t\in Q\}$